When connected across a battery, a capacitor stores a charge of 50...

Initial charge on the capacitor is given by $Q_i = CV = 50 \mu\text{C}$.

When a dielectric of dielectric constant $K$ is inserted, the new capacitance becomes $C' = KC$.

Since the battery remains connected, the potential difference $V$ across the capacitor remains constant.

Final charge on the capacitor is $Q_f = C'V = KCV = K Q_i$.

The additional charge that flows through the battery is $\Delta Q = Q_f - Q_i = 100 \mu\text{C}$.

Therefore, the final charge is $Q_f = 50 + 100 = 150 \mu\text{C}$.

Using $Q_f = K Q_i$, we get $150 = K \times 50$.

Solving for $K$, we get $K = 3$.