The growth of current in an LR circuit is governed by the equation: $i = i_0 (1 - e^{-t/\tau})$

1. Set the Condition: We need to find the time $t$ when the current $i$ reaches 50% of its steady-state value $i_0$. $0.5 i_0 = i_0 (1 - e^{-t/\tau})$

2. Simplify: $0.5 = 1 - e^{-t/\tau}$ $e^{-t/\tau} = 1 - 0.5 = 0.5$

3. Solve for t: Taking the natural logarithm (ln) on both sides: $-t/\tau = \ln(0.5)$ $-t/\tau = -\ln(2)$ $t = \tau \ln(2)$

4. Calculation: Given $\ln(2) = 0.693$, we have: $t = 0.693 \tau$

(Note: The provided question text is missing the specific value of the time constant $\tau$. However, for the answer to be 6.93 ms, the time constant $\tau$ must be 10 ms. Assuming $\tau = 10 \text{ ms}$, then $t = 0.693 \times 10 \text{ ms} = 6.93 \text{ ms}$.)