Answer
The area of the hole increases by $~3.80~\times 10^{-10}~m^2$
Work Step by Step
We can find the increase in area of the hole when the brass plate is heated:
$\Delta A = 2~\alpha~\Delta T~A$
$\Delta A = (2)(19\times 10^{-6}~K^{-1})(10.0~K)~(1.00\times 10^{-6}~m^2)$
$\Delta A = 3.80~\times 10^{-10}~m^2$
The area of the hole increases by $~3.80~\times 10^{-10}~m^2$
