To find the current in the secondary coil of the transformer, we first need to calculate the output power, taking into account the efficiency. The efficiency ($$ \eta $$) of the transformer is given by the ratio of the output power ($$ P_{\text{out}} $$) to the input power ($$ P_{\text{in}} $$) times 100%.

The given efficiency is $$ \eta = 80\% $$ or $$ \eta = 0.8 $$ in decimal form. The input power is also given as $$ P_{\text{in}} = 4 \text{kW} $$ or $$ P_{\text{in}} = 4000 \text{W} $$.

Let's calculate $$ P_{\text{out}} $$ using the efficiency formula:

$$ P_{\text{out}} = \eta \times P_{\text{in}} = 0.8 \times 4000 \text{W} = 3200 \text{W} $$

Now that we have $$ P_{\text{out}} $$, we can calculate the secondary current ($$ I_{\text{secondary}} $$) using the formula:

$$ P_{\text{out}} = V_{\text{secondary}} \times I_{\text{secondary}} $$

We are given $$ V_{\text{secondary}} = 240 \text{V} $$.

Isolating $$ I_{\text{secondary}} $$ gives us:

$$ I_{\text{secondary}} = \frac{P_{\text{out}}}{V_{\text{secondary}}} $$

Substituting the known values:

$$ I_{\text{secondary}} = \frac{3200 \text{W}}{240 \text{V}} $$

$$ I_{\text{secondary}} = \frac{3200}{240} $$

$$ I_{\text{secondary}} = 13.33 \text{A} $$

Therefore, the current in the secondary coil is $$ 13.33 \text{A} $$, which corresponds to Option B.