First compute the second moment:
[tex]\mathbb E[X^2]=\displaystyle\sum_x x^2p(x)=13^2\cdot0.07+\cdots+31^2\cdot0.08\approx447.13[/tex]
Then the variance of [tex]X[/tex] is
[tex]\mathbb V[X]=\mathbb E[X^2]-\mathbb E[X]^2\approx447.13-20.59^2\approx23.18[/tex]