Use the silicon diode model

For a normal silicon diode, when it is forward biased, the voltage drop across it is approximately $0.7\text{ V}.$

Identify conducting path

Since the positive terminal of the $5\text{ V}$ battery is connected to point $X$, the diode arrangement makes the relevant diode(s) forward biased in the current path from $X$ to $Y$.

In the given circuit, the effective forward-biased path includes one silicon diode in series with the resistor combination seen between $X$ and $Y$.

Therefore, the net voltage available across the resistor part is $V_R = 5 - 0.7 = 4.3\text{ V}.$

Effective resistance

From the circuit, the equivalent resistance in the conducting path is $R_{\text{eq}} = 10\,\Omega.$

Calculate current supplied by battery

Using Ohm’s law, $I = \frac{V_R}{R_{\text{eq}}} = \frac{4.3}{10} = 0.43\text{ A}.$

Match with options

$I \approx 0.43\text{ A}$ which corresponds to Option C.

Comparison with stored answer

Stored correct answer: C

My derived answer: C

Hence, they agree.