Answer
Please see the work below.
Work Step by Step
We know that:
$Molar\space mass\space of \space SnO_2=118.7+2(16.00)=150.7g$
Now:
$\%Sn=\frac{118.7}{150.7}\times 100\%=78.77\%$
and $\%$ composition of oxygen is calculated as follows
$\%O=\frac{2(16.00)}{150.7}\times 100\%=21.23\%$
