b
Centre of mass of the rod is given by:

\(\begin{array}{l}
{x_{cm}} = \frac{{\int\limits_0^L {(ax + \frac{{b{x^2}}}{L})dx} }}{{\int\limits_0^L {\left( {a + \frac{{bx}}{L}} \right)dx} }}\\
\,\,\,\,\,\,\,\,\, = \frac{{\frac{{a{L^2}}}{2} + \frac{{b{L^2}}}{3}}}{{aL + \frac{{bL}}{2}}} = \frac{{L\left( {\frac{a}{2} + \frac{b}{3}} \right)}}{{a + \frac{b}{2}}}\\
Now\,\frac{7}{{12}} = \frac{{\frac{a}{2} + \frac{b}{3}}}{{a + \frac{b}{2}}}
\end{array}\)

On solving we get, \(b = 2a\)