Hello Nano,
Newton's law of Cooling states that the time rate of change of the temperature $T$ of an object is proportional to the difference between the ambient temperature $M$ and the temperature of the object. Stated mathematically, this is:
$$\frac{dT}{dt}=-k(T-M)$$ where $$T(0)=T_0,\,0<k\in\mathbb{R}$$ and $$T>M$$.
The ODE is separable and may be written:
$$\frac{1}{T-M}\,dT=-k\,dt$$
Integrating, using the boundaries, and dummy variables of integration, we find:
$$\int_{T_0}^{T(t)}\frac{1}{u-M}\,du=-k\int_0^t v\,dv$$
$$\ln\left(\frac{T(t)-M}{T_0-M} \right)=-kt$$
If we know another point $$\left(t_1,T_1 \right)$$ we may now find $k$:
$$-k=\frac{1}{t_1}\ln\left(\frac{T_1-M}{T_0-M} \right)$$
Hence, we find:
$$t=\frac{t_1\ln\left(\frac{T(t)-M}{T_0-M} \right)}{\ln\left(\frac{T_1-M}{T_0-M} \right)}$$
Using the following given data:
$$t_1=10,\,T(t)=155,\,M=70,\,T_0=800,\,T_1=700$$
we find:
$$t=\frac{10\ln\left(\frac{155-70}{800-70} \right)}{\ln\left(\frac{700-70}{800-70} \right)}=\frac{10\ln\left(\frac{17}{146} \right)}{\ln\left(\frac{63}{73} \right)}\approx145.9628332694268$$
Therefore d) is the correct answer.