Answer
$R_{eq} =24\Omega$
Work Step by Step
let $R_1$ be the equivalent resistance of $60\Omega $, $ 60\Omega$ and $45\Omega $ which are in parallel
$\frac{1}{R_1 } =\frac{1}{60\Omega }+\frac{1}{60\Omega }+\frac{1}{45\Omega}$
$R_1 = 18\Omega$
now $18\Omega $ is in series with $42\Omega$ let $R_2$ be their equivalent
$R_2 =18\Omega + 42\Omega = 60\Omega$
$R_2$ and $40\Omega $ are in parallel
$\therefore \frac{1}{R_{eq} } =\frac{1}{ 60\Omega}+\frac{1}{40\Omega }$
$R_{eq} =24\Omega$
