Given : ‌‌R1=2Ω;
Let l1=l and a1=‌′a '
∵ The wire is stretched three time, so the new length (l2)=3l.
Let new area of crossection be a2
∵‌‌l1×a1=l2×a2
(Volume of wire is constant)
‌∴‌l×a‌=3l×a2
⇒‌a2=‌

l×a

3l

=‌

a

3

We know, R=‌

ρl

a

∴‌‌

R2

R1

=‌

l2

l1

×‌

a1

a2

(∵ρ=‌ constant ‌)
⇒‌

R2

2

=‌

3l

l

×‌

a

a∕3

⇒‌

R2

2

=3×3
⇒R2=18Ω