A solid rectangular brick is to be made from 1 cubic feet of clay. The length of the brick must be 3 times as long as its width. The width of brick for which it will have minimum surface area is  then $a,$ is $a^3$

Let width of a rectangular brick is  x
length of a rectangular brick is  3x
height of a rectangular brick is  h

$\begin{array}{l}\text{ Volume of the rectangular brick, }\left(x\right)\left(3x\right)\left(h\right)=1\\ \Rightarrow 3x^{2}h=1\\ \Rightarrow h=\frac{1}{3x^2}\\ \end{array}$

$\text{ Surface area of the brick }S=\left(x\right)\left(3x\right)+\left(3x\right)\left(h\right)+\left(h\right)\left(x\right)$

$\begin{array}{l}\Rightarrow S=6x^2+8hx\\ \Rightarrow S\left(x\right)=6x^2+8x\left(\frac{1}{3x^2}\right)\\ \Rightarrow S=6x^2+\frac{8}{3x}\\ \Rightarrow S^{\mathrm{\prime }}\left(x\right)=12x-\frac{8}{3x^2}\end{array}$

$\text{ For maxima or minima }{S}^{\mathrm{\prime }}\left(x\right)=0$

$\begin{array}{l}\Rightarrow 12x-\frac{8}{3x^2}=0\\ \Rightarrow x^3=\frac{2}{9}\\ \Rightarrow x={\left(\frac{2}{9}\right)}^{\frac{1}{3}}\\ S^{\prime \prime }\left({\left(\frac{2}{9}\right)}^{\frac{1}{3}}\right)=12+\frac{8}{\frac{2}{9}}=48>0\\ S^{\prime \prime }\left(x\right)=12+\frac{8}{x^3}\end{array}$

$\begin{array}{l}\therefore \text{ surface area is minimum when width }x={\left(\frac{2}{9}\right)}^{\frac{1}{3}}=a\\ \Rightarrow a^3=\frac{2}{9}\end{array}$

$Therefore, the correct answer is \left(2\right).$