Explanation

(b)
$Area = \int_{0}^{16/m^{2}}{\left( \sqrt{16x} - mx \right)dx = \frac{2}{3}}$
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$\Longrightarrow \ \left\lbrack 4 \bullet \frac{2}{3}x^{3/2} - \frac{{mx}^{2}}{2} \right\rbrack_{0}^{16/m^{2}} = \frac{2}{3}$
$\Longrightarrow \frac{1}{m^{3}}\left\lbrack \frac{512}{3} - \frac{256}{2} \right\rbrack = \frac{2}{3}$
$\Longrightarrow \ m^{3} = \frac{128}{3} \times \frac{3}{2} = 64$
$\Longrightarrow \ \ m = 4$