How can I get precisely a certain cubic cm by changing the following factors?
To avoid getting caught-up in specific numbers ...
Suppose you have $$a\cdot b \cdot c = d$$ but you want $d$ to become $e$. You can make this happen by multiplying both sides by $e/d$: $$\left(a\cdot b \cdot c \right)\cdot \frac{e}{d} = d\cdot\frac{e}{d} = e$$
Now, you can use the left-hand side's factor of $e/d$ to make adjustments to $a$, $b$, and/or $c$. If you just wanted to adjust one factor, you could write, say,
$$\left( a\cdot \frac{e}{d}\right)\cdot b\cdot c \;=\; e \tag{1}$$
If you wanted to adjust two factors proportionally (as is specifically requested in the question), you can "split" $e/d$ equally across the factors using a square root:
$$\frac{e}{d} = \sqrt{\frac{e}{d}}\cdot\sqrt{\frac{e}{d}} \qquad\to\qquad\left(a\cdot \sqrt{\frac{e}{d}}\right)\cdot\left(b\cdot \sqrt{\frac{e}{d}}\right)\cdot c \;=\; e \tag{2}$$
Finally, if you later decide you actually want to adjust your entire box proportionally, you can use cube roots:
$$\left(a\cdot\sqrt[3]\frac{e}{d}\right)\cdot\left(b\cdot\sqrt[3]\frac{e}{d}\right)\cdot \left(c\cdot\sqrt[3]\frac{e}{d}\right) \;=\; e \tag{3}$$
Naturally, the same type of thing works with any number of overall factors and desired adjustments, using higher-level roots as needed.
You have $3.833 \times 3.833 \times5.174=76.02 .$
You can change it by multiplying both sides by $\displaystyle \frac{76.04}{76.02}$.
We then have $3.833 \times 3.833 \times5.174 \times \displaystyle \frac{76.04}{76.02}=76.02 \times \displaystyle \frac{76.04}{76.02}$.
This comes out to $3.833 \times 3.833 \times5.17564=76.04 $
Actually $3.833 \cdot 3.833 \cdot 5.174 = 76.0158$ so the added volume will be $.0242$
One way is to think of it as adding a sheet $3.833 \cdot 3.833$ with a volume of $0.0242\ \text{cm}^3$. How thick does it have to be to equal that volume?
Hence, $\frac{0.0242}{3.833^2} = .00165$
So the dimensions will be $3.833 \cdot 3.833 \cdot (5.174 + .00165)$
$3.833 \cdot 3.833 \cdot 5.17565 = 76.040$