A 2 A current enters at the corner d of a square frame abcd having...

The total current entering the square frame at corner $d$ is $I = 2 \text{ A}$.

Since the frame is a square made of uniform wire, the resistance of the path $d \to a \to b$ is equal to the resistance of the path $d \to c \to b$. Therefore, the current splits equally into the two branches.

Current in each side of the frame is:
$I_{da} = I_{ab} = I_{dc} = I_{cb} = \dfrac{2}{2} = 1 \text{ A}$

The magnetic field is uniform, directed perpendicular to the plane of the frame and outwards (indicated by dots), so $\vec{B} = 0.1 \hat{k} \text{ T}$.

The magnitude of the magnetic force on each wire of length $L = 20 \text{ cm} = 0.2 \text{ m}$ is given by:
$F = I L B \sin(90^{\circ}) = 1 \times 0.2 \times 0.1 = 0.02 \text{ N}$

The direction of the force on each wire can be found using $\vec{F} = I(\vec{L} \times \vec{B})$:
For wire $da$: Current is downwards ($-\hat{j}$). $\vec{F}_{da} = 1(-0.2\hat{j} \times 0.1\hat{k}) = -0.02\hat{i} \text{ N}$ (towards left).
For wire $cb$: Current is downwards ($-\hat{j}$). $\vec{F}_{cb} = 1(-0.2\hat{j} \times 0.1\hat{k}) = -0.02\hat{i} \text{ N}$ (towards left).
For wire $dc$: Current is rightwards ($\hat{i}$). $\vec{F}_{dc} = 1(0.2\hat{i} \times 0.1\hat{k}) = -0.02\hat{j} \text{ N}$ (downward).
For wire $ab$: Current is rightwards ($\hat{i}$). $\vec{F}_{ab} = 1(0.2\hat{i} \times 0.1\hat{k}) = -0.02\hat{j} \text{ N}$ (downward).

Thus, the force is $0.02 \text{ N}$ on each wire; on $da$ and $cb$ towards left, and on $dc$ and $ab$ downward.