Answer
$4.50~\Omega$
Work Step by Step
We can find the equivalent resistance $R_{12}$ of $R_1$ and $R_2$ which are in parallel:
$\frac{1}{R_{12}} = \frac{1}{R_1} + \frac{1}{R_2}$
$\frac{1}{R_{12}} = \frac{1}{4.00~\Omega} + \frac{1}{4.00~\Omega}$
$\frac{1}{R_{12}} = \frac{2}{4.00~\Omega}$
$R_{12} = 2.00~\Omega$
This resistance is in series with $R_3$.
We can find the equivalent resistance:
$R_{eq} = R_{12}+R_3 = (2.00~\Omega)+(2.50~\Omega) = 4.50~\Omega$
