The ratio of Earth’s orbital angular momentum (about the sun) to its mass is  $4.4\times {10}^{15}{\text{\hspace{0.17em}m}}^{\text{2}}/\text{s}$. The area enclosed by the earth’s orbit is  $N\times {10}^{23}{m}^2$. Find the value of N ? 

Given
Angular momentum per mass $\frac{L}{M}=4.4\times {10}^{15}{\text{\hspace{0.17em}m}}^{\text{2}}/\text{s}$
By Kepler’s second law we use 
Areal velocity =  $\frac{\text{Area Swept}}{\text{Time for one Revolution of Earth about the sun }}$
Also we use Areal velocity =  $\frac{L}{2M}$
$\Rightarrow$ Area Swept =  $\left(\frac{L}{2M}\right)$ (Time for one Revolution of Earth about the sun) 
$\Rightarrow$ Area Swept =  $\frac{1}{2}\left(4.4\times {10}^{15}\right)\left(365\times 24\times 60\times 60\right)$
$\Rightarrow$ Area Swept =  $0.69\times {10}^{23}{\text{\hspace{0.17em}m}}^{\text{2}}$