The current passing through the battery in the given circuit, is:

To find the current passing through the battery, we can simplify the given circuit by analyzing the resistor network between nodes B and E.

Step 1: Simplify the connections in the circuit
By looking at the circuit diagram:
- Nodes A and F are connected directly by a zero-resistance wire, so they are at the same electrical potential ($V_A=V_F$).
- Nodes C and D are also connected directly by a zero-resistance wire, so they are at the same electrical potential ($V_C=V_D$).

Step 2: Identify the Wheatstone Bridge
Let us analyze the resistor network between the terminals B and E:
- Resistor between B and A: $R_{\mathrm{AB}}=5\Omega$
- Resistor between B and D (via C): $R_{\mathrm{BD}}=2.5\Omega$
- Resistor between A (via F) and E: $R_{\mathrm{AE}}=3\Omega$
- Resistor between E and D: $R_{\mathrm{ED}}=1.5\Omega$
- Diagonal resistor between A and D: $R_{\mathrm{AD}}=6\Omega$

Let's check the ratio of the opposite arms of the bridge:
$\frac{R_{\mathrm{AB}}}{R_{\mathrm{AE}}}=\frac{5}{3}$
$\frac{R_{\mathrm{BD}}}{R_{\mathrm{ED}}}=\frac{2.5}{1.5}=\frac{5}{3}$

Since the ratio is equal:
$\frac{R_{\mathrm{AB}}}{R_{\mathrm{AE}}}=\frac{R_{\mathrm{BD}}}{R_{\mathrm{ED}}}$
This forms a balanced Wheatstone bridge. Consequently, the potential at node A is equal to the potential at node D ($V_A=V_D$). No current flows through the diagonal $6\Omega$ resistor, and it can be removed from the calculation.

Step 3: Calculate the equivalent resistance of the bridge ($R_{\mathrm{BE}}$)
With the $6\Omega$ resistor removed, the circuit consists of two parallel branches connected between B and E:
- Upper branch: $R_{\mathrm{AB}}+R_{\mathrm{AE}}=5\Omega +3\Omega =8\Omega$
- Lower branch: $R_{\mathrm{BD}}+R_{\mathrm{ED}}=2.5\Omega +1.5\Omega =4\Omega$

The equivalent resistance $R_{\mathrm{BE}}$ is:
$R_{\mathrm{BE}}=\frac{8\times 4}{8+4}=\frac{32}{12}=\frac{8}{3}\Omega$

Step 4: Calculate the total resistance of the circuit ($R_{\mathrm{total}}$)
The battery of $5V$ is connected in series with the bridge network ($R_{\mathrm{BE}}$), the upper external resistors, and the lower external resistor:
- Upper external resistance: $1.5\Omega +5.5\Omega =7\Omega$
- Lower external resistance: $\frac{1}{3}\Omega$
- Bridge equivalent resistance: $\frac{8}{3}\Omega$

Thus, the total resistance is:
$R_{\mathrm{total}}=7+\frac{8}{3}+\frac{1}{3}=7+\frac{9}{3}=7+3=10\Omega$

Step 5: Calculate the current passing through the battery ($I$)
Using Ohm's law:
$I=\frac{V}{R_{\mathrm{total}}}=\frac{5V}{10\Omega }=0.5A$

The correct answer is 0.5 A.