Let the resistance of the wire that BREAKS be r .

Let the equivalent resistance be R and the second resistor that is joined with 'r' be r₁ .

As the resistors are connected in parallel :

1/R = 1/r + 1/r₁

= > 1/(6/5) = 1/r + 1/r₁

= > 5/6 = 1/r + 1/r₁

= > 1/r = 5/6 - 1/r₁ .

If one of the resistance breaks then the effective resistance breaks .

The resistance of 'r' is broken and hence there is only one resistor connected in the CIRCUIT .

The resistance of the wire becomes 2 ohms which tells us that the other resistor has a resistance of 2 ohms .

Putting r₁ = 2 in that EQUATION , we get :

5 / 6 Ω = 1 / r + 1 / 2 Ω

= > 5 / 6 Ω = ( 2 + r ) / 2 r

= > 10 r = 12 Ω + 6 r

= > 10 r - 6 r = 12 Ω

= > 4 r = 12 Ω

= > r = 12/4 Ω

= > r = 3 Ω

The resistance of the wire that break is 3 Ω .

NOTE :

When the resistors are in parallel , the equivalent resistance denoted by R will be R = 1/R₁ + 1/R₂ where R₁ and R₂ are the resistances connected in parallel .

When the resistors are in series , the equivalent resistance denoted by R will be R = R₁ + R₂ where R₁ and R₂ are the resistances connected in series .