Answer
Please see the work below.
Work Step by Step
We can find the required combination as follows:
The parallel combination of resistors $R_1$ and $R_2$ gives
$R^{\prime}=\frac{R_1\times R_2}{R_1+R_2}$
$\implies R^{\prime}=\frac{220\Omega\times 79\Omega}{220\Omega+79\Omega}$
$R^{\prime}=58\Omega$
Now $R^{\prime}$ is connected in series with $R_3$ to give
$R_{eq}=R^{\prime}+R_3$
We plug in the known values to obtain:
$R_{eq}=58\Omega+92\Omega$
$R_{eq}=150\Omega$
Thus, the required connections should be $R_1||R_2$ and $R_3$ in series.
