$$t_{1/4}$$ can be taken as the time taken for the concentration of a reactant to drop to $$\frac{3}{4}$$ of its initial value. If the rate constant for a first order reaction is $$K$$, the $$t_{1/4}$$ can be written as

For a first-order reaction, the instantaneous concentration $$[A]$$ varies with time $$t$$ according to the integrated rate law

$$[A] = [A]_0 \, e^{-K t}$$

where $$[A]_0$$ is the initial concentration and $$K$$ is the first-order rate constant.

By definition, $$t_{1/4}$$ is the time taken for the concentration to fall to $$\tfrac{3}{4}$$ of its initial value, that is

$$[A] = \frac{3}{4}[A]_0$$ at $$t = t_{1/4}$$.

Substituting this condition into the rate law gives

$$\frac{3}{4}[A]_0 = [A]_0 \, e^{-K t_{1/4}}.$$

Cancelling $$[A]_0$$ from both sides, we get

$$\frac{3}{4} = e^{-K t_{1/4}}.$$

Taking the natural logarithm of both sides:

$$\ln\!\left(\frac{3}{4}\right) = -K t_{1/4}.$$

Hence

$$t_{1/4} = -\frac{1}{K}\,\ln\!\left(\frac{3}{4}\right).$$

Now, $$\ln\!\left(\frac{3}{4}\right) = \ln 3 - \ln 4 = 1.0986 - 1.3863 = -0.2877$$ (to four significant figures).

Therefore

$$t_{1/4} = \frac{0.2877}{K} \approx \frac{0.29}{K}.$$

Option B which is: $$0.29/K$$