For a first order reaction, A $$\rightarrow$$ P, t$$_{1/2}$$ (half-life) is 10 days. The time required for $$\frac{1}{4}$$th conversion of A (in days) is: (ln 2 = 0.693, ln 3 = 1.1)

For the first-order reaction $$A \rightarrow P$$, we first recall the two fundamental relations that connect rate constant $$k$$, half-life $$t_{1/2}$$ and concentration:

1. The half-life formula for a first-order reaction is stated as $$t_{1/2}=\frac{0.693}{k}.$$

2. The integrated rate law is $$\ln\!\left(\frac{[A]_0}{[A]}\right)=kt,$$ where $$[A]_0$$ is the initial concentration and $$[A]$$ is the concentration at time $$t$$.

We are given the half-life $$t_{1/2}=10\ \text{days}$$. Using the half-life formula, we substitute and solve for $$k$$:

$$10=\frac{0.693}{k} \quad\Rightarrow\quad k=\frac{0.693}{10}=0.0693\ \text{day}^{-1}.$$

Next, “$$\dfrac14$$-th conversion” means that one-quarter of the reactant has reacted, so three-quarters remains. Mathematically,

$$\frac{[A]}{[A]_0}=\frac34 \quad\Longrightarrow\quad \frac{[A]_0}{[A]}=\frac43.$$

We now insert this ratio into the integrated rate law:

$$\ln\!\left(\frac43\right)=kt.$$

Substituting the value of $$k$$ obtained above gives

$$t=\frac{\ln\!\left(\tfrac43\right)}{0.0693}.$$

To evaluate the natural logarithm, we expand it using the property $$\ln\!\left(\tfrac43\right)=\ln4-\ln3.$$ Since $$\ln2=0.693,$$ we first find $$\ln4$$:

$$\ln4=\ln(2^2)=2\ln2=2(0.693)=1.386.$$

The question supplies $$\ln3=1.1,$$ hence

$$\ln\!\left(\tfrac43\right)=1.386-1.1=0.286.$$

We now divide by the rate constant:

$$t=\frac{0.286}{0.0693}\ \text{days}\approx4.13\ \text{days}.$$

Rounding suitably, the time required is about $$4.1\ \text{days}.$$

Hence, the correct answer is Option C.