Answer
a) $I=\frac{\mathcal{E}}{R}$
b) $R_{eq}=R$
Work Step by Step
a) Number the resistors 1 to 5 from left to right.
$I=I_1+I_2$
$I=I_4+I_5$
$I_1=I_3+I_4$
$I_2+I_3=I_5$
$I_1R+I_3R-I_2R=0$
$I_2=I_1+I_3$
$-I_4R+I_5R+I_3R=0$
$I_4=I_3+I_5$
$I_1R+I_3R+I_5R=\mathcal{E}$
$I_1+I_3+I_5=\frac{\mathcal{E}}{R}$
$I_1+I_2+0-I_4-I_5=0$
$I_1+0-I_3-I_4+0=0$
$0+I_2+I_3+0-I_5=0$
$I_1-I_2+I_3+0+0=0$
$I_1+0+I_3+0+I_5=\frac{\mathcal{E}}{R}$
We can solve this by using a 5 by 5 matrix.
The results are as follows:
$I=\frac{\mathcal{E}}{R}$
b) From Ohm's Law: $\mathcal{E}=IR_{eq}=(\frac{\mathcal{E}}{R})R_{eq}$
$R_{eq}=R$