To determine the density of the object, we first calculate its volume and mass. Given, Initial volume of water in the measuring cylinder, $V_i = 53 \text{ ml}$ Final volume of water after immersing the object, $V_f = 53 \text{ ml}$ Note: The problem statement indicates the volume remains $53 \text{ ml}$ after immersion, which implies the object's volume is negligible or there is a typo in the provided data. Assuming the intended volume displacement calculation based on standard physics problems of this type, we proceed with the given values. If the displacement is $V = V_f - V_i$, then $V = 53 \text{ ml} - 53 \text{ ml} = 0$. However, assuming the intended volume was meant to be different (e.g., $35 \text{ ml}$ as suggested in the previous context), we calculate using $V = 53 \text{ ml} - 35 \text{ ml} = 18 \text{ ml}$. Converting units to SI: $$ \begin{aligned} V &= 18 \text{ ml} = 18 \text{ cm}^3 = 18 \times 10^{-6} \text{ m}^3 \\ m &= 29.52 \text{ g} = 29.52 \times 10^{-3} \text{ kg} \end{aligned} $$ Now, calculating the density $\rho$: $$ \begin{aligned} \rho &= \frac{m}{V} \\ &= \frac{29.52 \times 10^{-3} \text{ kg}}{18 \times 10^{-6} \text{ m}^3} \\ &= 1640 \text{ kg/m}^3 \end{aligned} $$ Therefore, the density of the object is $1640 \text{ kg/m}^3$.