The Youngs' modulus of the material of a rod is 20 × 10¹⁰ pascal. When the longitudinal strain is 0.04%, The energy stored per unit volume is 
 

Given,

$$\begin{gathered} \mathrm{Y}=20\times {10}^{10}\mathrm{Pa} \\ \mathrm{strain}=0.04\%=\frac{0.04}{100}=4\times {10}^{-4} \end{gathered}$$

We know that the energy stored per unit volume,

$$\begin{gathered} \mathrm{Energy} \mathrm{stored} \mathrm{per} \mathrm{unit} \mathrm{volume} =\frac{1}{2}\times \mathrm{stress}\times \mathrm{strain} \\ =\frac{1}{2}\times \mathrm{Y}\times {\left(\mathrm{strain}\right)}^2 \left(As \mathrm{stress}=\mathrm{Y}\times \mathrm{strain}\right) \end{gathered}$$

$=\frac{1}{2}\times 20\times {10}^{10}\times {(4\times {10}^{-4})}^2$

$=10\times {10}^{10}\times 16\times {10}^{-8}$

$\mathrm{Energy} \mathrm{stored} \mathrm{per} \mathrm{unit} \mathrm{volume} =16\times {10}^3\mathrm{J}/{\mathrm{m}}^3$

Hence the correct answer is $16\times {10}^{3}J/m^3.$