How many grams of solid sodium bicarbonate will be consumed?
$$\ce{Cu(NO3)2_{(aq)} + 2NaHCO3_{(s)} -&gt; CuCO3_{(s)} + 2NaNO3_{(aq)} + H2O_{(l)} + CO2_{(g)}}$$
The equation is telling us that for every $1$ mole of $\ce{Cu(NO3)2}$ consumed, there are $2$ moles of $\ce{NaHCO3}$ correspondingly consumed in the reaction.

So, let us first find how many moles of $\ce{Cu(NO3)2}$ is consumed. Here,
$$\text{Molarity} = \frac{\text{#Moles of } \ce{Cu(NO3)2}}{\text{Volume of Solution}}\\ \implies \text{#Moles of } \ce{Cu(NO3)2} = 0.240 \text{ mol.L}^{-1}\times 510  \text{ mL} \times \frac{1\text{ L}}{1000 \text{ mL}} = 0.1224 \text{ mol}$$

Now, think.

$$1 \text{ mole } \ce{Cu(NO3)2}  \iff 2 \text{ moles } \ce{NaHCO3}\\
0.1224 \text{ moles } \ce{Cu(NO3)2} \iff 0.2448 \text{ moles } \ce{NaHCO3}$$

$\text{Weight of } \ce{NaHCO3}= 0.2448 \text{ mol} \times 84.007 \text{ g mol}^{-1} = 20.5649136 \text{ g}$