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A common misconception is that kriging estimates may be simply exponentiated to recover the field values.

Sebastien Rochette's suggests a back-transformation for field values y following Laurent (1963):

Because the prediction of log(y) is based on a Gaussian distribution, in many cases an additional correction factor is needed because the expected value of back-transformed lognormal kriging estimates are biased--not equal to the sample mean. One method is to multiply the above equation by a correction factor: the ratio of the sample mean to the back-transformed means (k_o in the equation below).

Here is a paper on the issue of back-transforming kriging estimates. It draws on the work of (Journel, 1978).

And here is a fully reproducible example in R:

# load required packages, and download missing ones
if (!require("pacman")) install.packages("pacman")
pacman::p_load(sp, raster, gstat, ggplot2, magrittr)

data(meuse) # load the meuse data set

# convert meuse into a spatial object
# first, prepare the 3 components: coordinates, data, and proj4string
coords <- meuse[ , c("x", "y")]   # coordinates
data   <- meuse[ , 3:14]          # data
crs    <- CRS("+init=epsg:28992") # proj4string of coords

# make the spatial points data frame object
d <- SpatialPointsDataFrame(coords      = coords,
                            data        = data, 
                            proj4string = crs)

r <- raster(d)                       # create raster to interpolate over
res(r) <- 100                        # raster resolution (100 meters)
g <- as(r, "SpatialGrid")            # convert raster to spatial grid object


d$zinc <- log(d$zinc)                # log transform field values
  
gs <- gstat(formula = zinc ~ 1,      # spatial data, so fitting xy as idp vars
            locations = d)           # spatial object
  
v <- variogram(gs,                   # gstat object
               width = 25)           # lag distance
  
# plot the variogram
# plot(v)                              

fve <- fit.variogram(v,              # takes `gstatVariogram` object
                     vgm(0.6,        # partial sill: semivariance at the range
                         "Exp",      # linear model type
                         1000,       # range: distance where model first flattens out
                         0.01))      # nugget
  
# plot variogram and fit
plot(v, fve)

# ordinary kriging 
kp <- krige(zinc ~ 1, d, g, model = fve)
  
# backtransformed
bt <- exp( kp@data$var1.pred + (kp@data$var1.var / 2) )
  
# means of backtransformed values and the sampled values
mu_bt <- mean(bt)
mu_original <- mean(exp(d$zinc))
  
# these means differ...  
> mu_bt
[1] 673.0071
> mu_original
[1] 469.7161

# ...thus make another correction to remove kriging bias in sample mean
btt <- bt * (mu_original/mu_bt)             # correct backtransfomed vals
kp@data$var1.pred <- btt                    # overwrite w/ correct vals 
names(kp) <- c("Prediction", "variance")
spplot(kp, "Prediction", main = "Zinc concentration")

# we can view the motivation for this transformation by comparing the 
# original zinc values to the kirging estimates generated by the
# backtransformation, and the corrected backtransformation
rbind.data.frame(
  data.frame(val = exp(d$zinc), class = "Original Values"), 
  data.frame(val = bt,          class = "Back-trans"), 
  data.frame(val = btt,         class = "Corrected Back-trans")
) %>% 
  ggplot(aes(val)) + 
  geom_density(aes(fill=class), alpha = 0.3) +
  facet_wrap(~class, ncol=1) +
  guides(fill = FALSE)

Summary

It's clear that the density distribution of the corrected backtransform aligns with the sample density distribution, indicating less bias compared to the simple backtransformation without the correction coefficient.