Answer
Area $A=3.17460\times10^{2}m^2$
Work Step by Step
From table number 11.1
Density of silver $\rho=10500kg/m^3$
from $\rho=\frac{M}{V}$
$V=\frac{M}{\rho}$
given that mass used for pounding $M=1kg$
so volume of $1kg $ silver will be $V=\frac{1kg}{10500kg/m^3}$
$V=9.52381\times10^{-5}m^3$
now this is converted to thin sheets of thickness $t=3.00\times10^{-7}m$
if area of such sheet is $A$
then volume of sheet is $V$=$thickness$ x $area$
$V=t\times A$
$A=\frac{V}{A}$
putting $V=9.52381\times10^{-5}m^3$, $t=3.00\times10^{-7}m$
Area $A=\frac{9.52381\times10^{-5}m^3}{3.00\times10^{-7}m}$
Area $A=3.17460\times10^{2}m^2$
