Using Kirchhoff's Current Law, determine the value of current I in the given circuit diagram.

Two currents, 3 A and 1.5 A, are entering this junction. Let the current leaving this junction be $I_1$.

$$I_1 = 3 \text{ A} + 1.5 \text{ A}$$ $$I_1 = 4.5 \text{ A}$$

Current $I_1$ (4.5 A) is entering this junction. Current 0.5 A is leaving, and another current, let's call it $I_2$, is also leaving.

$$I_1 = 0.5 \text{ A} + I_2$$ $$4.5 \text{ A} = 0.5 \text{ A} + I_2$$ $$I_2 = 4.5 \text{ A} - 0.5 \text{ A}$$ $$I_2 = 4.0 \text{ A}$$

Current $I_2$ (4.0 A) and 2.5 A are entering this junction. Let the current leaving this junction be $I_3$.

$$I_3 = I_2 + 2.5 \text{ A}$$ $$I_3 = 4.0 \text{ A} + 2.5 \text{ A}$$ $$I_3 = 6.5 \text{ A}$$

Current $I_3$ (6.5 A) is entering this junction. Currents $I$ and 1.5 A are leaving this junction.

$$I_3 = I + 1.5 \text{ A}$$ $$6.5 \text{ A} = I + 1.5 \text{ A}$$ $$I = 6.5 \text{ A} - 1.5 \text{ A}$$ $$I = 5.0 \text{ A}$$

The calculated value for $I$ is 5.0 A.
Comparing this with the given options:
[A] 0.3 A
[B] 0.5 A
[C] 1.3 A
[D] इनमें से कोई नहीं (None of these)

Since 5.0 A is not among options A, B, or C, the correct option is [D].

The final answer is $\text{D}$.