Solution:
The given Stoke's formula is,
F = 6πηrv ------------------- (i)
The dimension of F = [MLT-2]
The dimension of η = [ML-1T-1]
The dimension of r = [L]
The dimension of v = [LT-1]
Substituting the value of dimensions in equation (i), we get,
\(\left[MLT^{-2}\right]=\left[ML^{-1}T^1\right]\left[L\right]\left[LT^{-1}\right]=\left[MLT^{-2}\right]\)
Since, the dimensions on the left hand side are equal to the dimensions on the right hand side. So the formula is dimensionally correct.