From the figure, 6Ω and 3Ω are in parallel, so,
‌

1

Rp

‌=‌

1

6

+‌

1

3

=‌

1+2

6

=‌

3

6

=‌

1

2

∴‌‌RP‌=2Ω
Also, 4Ω and 12Ω are in parallel, so
‌

1

RP‌′

‌=‌

1

4

+‌

1

12

=‌

3+1

12

=‌

4

12

=‌

1

3

∴‌‌RP′‌=3Ω
Now, the given parallel combinations are in series. So we get,
Req=2+3=5Ω