 Solution

Both factors are assigned at orchard level. Therefore the experimental units are the twelve orchards; the trees are observational units within them. The six combinations form a balanced factorial design, replicated twice. The orchard ANOVA stratum has $12-1=11$ statistical degrees of freedom. Spray uses $2-1=1$, pruning uses $3-1=2$, and their interaction term uses $(2-1)(3-1)=2$, leaving six for error.

Dividing each treatment sum of squares in ANOVA by its statistical degrees of freedom and using $1440/6=240$ as the denominator gives all missing entries:

The unrounded F-test statistics are $998/240=4.15833\dots$, $560/240=2.33333\dots$ and $202/240=0.841667\dots$. The within-orchard tree mean square in ANOVA, 180, is not the treatment error denominator: using it would confuse subsampling with independent replication. The tree ANOVA stratum has $12(30-1)=348$ statistical degrees of freedom; $1+11+348=360$ is the uncorrected total, and the corrected total is 359.