Give the $SD = 5^{\circ}C$
Variance $=(S.D)^2$
$\qquad= (5^{\circ}C)^2$$=25^{\circ}C$
To convert $^{\circ}C$ to $^{\circ}F$
=> $ \large\frac{^{\circ}C \times 9}{5} $$+32$
=> $\large\frac{25^{\circ} C \times 9}{5} $$+32$
=> $45 +32= 77^{\circ}F$
The nearest answer is 81
Hence A is the correct answer.