The rectangular box is a non ohmic resistance following $I = \frac{V ^{4}}{125}$ where I is current through it (inA) & V voltage across it (in V). At the balanced condition of wheatstone bridge.

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The rectangular box is a non ohmic resistance following I = 125 V 4 ​ where I is current through it (inA) & V voltage across it (in V). At the balanced condition of wheatstone bridge.

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At balance the bridge ratio gives the box resistance equal to the opposite $1 Ω$ arm, and the galvanometer carries no current, so the two arms split the total current equally.

For the box, balance requires $V = 5 V$ : then $I = \frac{V ^{4}}{125} = \frac{625}{125} = 5 A$ , consistent with $V / I = 1 Ω$ - (B) and (C) are correct.

The other 1 $Ω$ -1 $Ω$ arm carries the same 5 A, so the total current $i = 5 + 5 = 10 A$ - (A) is correct.

Each arm drops 5 V, so the source voltage is $V_{S o u rce} = 5 + 5 = 10 V$ - (D) is correct.

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