Concentration and Half Life of reactants A and B
The whole reaction isn't first order and that can be arrived at using one simple concept. The unit of rate constant $k$ is $\pu{M^{-1} s^{-1}}$ and not $\mathrm s^{-1}$.

Since reaction is first order with respect to both reactant A and B, I can write,
$$ k=\frac{1}{t} \ln{\frac{a}{a-x}} $$

This would have been correct if and only if the whole reaction was a first order reaction. Not a first order reaction with respect to each reactant individually. For example:
$$\ce{A-&gt;B}\quad \text{Rate}=k[A]$$
That is not the case here. Here the reaction would be better phrased as:
$$\ce{A + B-&gt;C}\quad \text{Rate} = k[A][B]$$
Which means that this is a second order reaction and not a first order reaction.
Now, the other issue is that this second order reaction isn't of the form $\ce{2A-&gt; B}$ where the rate would have been relatively easy to find using a generic law and hence we move to a more basic proof for a reaction $\ce{A + B -&gt; C}$. As given above, the rate law for this expression is &quot;Rate $= k[A][B]$&quot; Now we draw a RICE table for this generic reaction to get:
\begin{array}{|l|c|} \hline
\mathrm{R} &amp; \mathrm A &amp; \mathrm B &amp;\mathrm C \\ \hline
\mathrm I &amp; a &amp; b &amp; 0 \\
\mathrm C &amp; -x &amp; -x &amp; x \\
\mathrm E &amp; a-x &amp; b-x &amp; x \\ \hline
\end{array}
Hence, we can see that for a time t, the rate of reaction can be defined as:
$$ \text{Rate} = -\frac{\mathrm d[A]}{\mathrm dt}=k(a-x)(b-x)$$
Differentiating $[A]$ with respect to $t$, we get:
$$-\frac{\mathrm d[A]}{\mathrm dt} = \frac{\mathrm dx}{\mathrm dt}$$
Substituting this in the previous equation and integrating with respect to x, we get:
$$kt = \frac{1}{b-a}\ln \frac{a(b-x)}{b(a-x)}$$
Now, in the given question, we are asked to find the value of $[A]$ and $[B]$ after $\pu{100 s}$. Therefore, $t=\pu{100 s}$, $a = \pu{0.1 M}\,,b=\pu{6.93 M}$ and $k = \pu{2 \times 10^-3 M^{-1} s^{-1}}$.
Solve for x in this equation and find the value of $a-x$ and $b-x$.
This isn't a complete solution, but the rest is just manipulating the expression using mathematics and has no actual relation to kinetics.
A hint to make calculations simpler would arise from noticing that $a&lt;&lt;b$ and so $b-a$, $b-x$ can both be approximated to $b$.