Can you solve it in first attempt?

Aha! Missed It In First Attempt As I Gave the answer as 20.24% (which is actually the percentage of A In Compound Formed By A And C)

For The Sake Of Simplicity We Can Assume That Valencies Of All The Three

Elements Are 1 (You can take variable valencies as a,b,c etc but they are going

to cancel out in the end Because of the consequence of law of definite proportions) . so there is no need to mess your work .

Let The Molecular Masses Of Elements Be X, Y and Z Respectively .

Now Simply Plug In The Values In Data Provided

X/X+Y = 3/4

Y/Y+Z = 7.8/100

From Here We Can Get Values Of X And Z In Terms Of Y .

And What We Require is

Z/Z+X * 100

So we can simply substitute values of Z And X In The Above Equation

And On Calculation It Will Yield 79.8 (79.75899 to be precise)

Note - The Problem Might Be Attempted Using Laws of reciprocal proportions but it

will not yield the Correct Answer Because it is not applicable for Any three

pair of Elements . You can go on by law of equivalence as secondary method

of solving the problem

Let the molecular masses of $\ce{A}$ , $\ce{B}$ and $\ce{C}$ be $m_A$ , $m_B$ and $m_C$ respectively. Then, we have:

$\begin{cases} \dfrac{m_A}{m_B} = \dfrac{0.75}{0.25} & \Rightarrow m_A = 3 m_B \\ \dfrac{m_B}{m_C} = \dfrac{0.078}{0.922} & \Rightarrow m_C = 11.8205 m_B \end{cases}$

Then $x = \dfrac{m_C}{m_{AC}} = \dfrac{m_C}{m_A+m_C} = \dfrac {11.8205 m_B}{3 m_B + 11.8205 m_B} = 0.7976 \approx \boxed{79.8} \%$

This is pure mathematics for same valence in these compounds.

$\frac{A}{A + B} = 75$ % = $\frac34 \implies \frac{B}{A} = \frac43 -1 = \frac13$

$\frac{B}{B + C} = 7.8$ % = $\frac{39}{500} \implies \frac{C}{B} = \frac{500}{39} -1 = \frac{461}{39}$

With $\frac{C}{A} = \frac{461}{117} \implies \frac{A}{C} = \frac{117}{461},$

$\frac{C}{A + C} = \frac{1}{1 + \frac{A}{C}} = \frac{1}{1 + \frac{117}{461}} = \frac{461}{578} \times 100$ % = (79.75778546712802768166089965397+)%

Answer: $\boxed{79.8}$