Answer
$$15.811$$
Work Step by Step
$$\eqalign{
& {\text{We can note that }}\sqrt {250} {\text{ is a solution of the equation }}{x^2} - 250 = 0.{\text{ }} \cr
& {\text{Let }}f\left( x \right) = {x^2} - 250,\,\,\,\,\,\,f'\left( x \right) = 2x \cr
& {\text{We have that }}{\left( {15} \right)^2} = 225{\text{ and }}{\left( {16} \right)^2} = 256 \cr
& {\text{Since 15 < }}\sqrt {15} < 16,{\text{ we can use }}{c_1} = 15{\text{ as the first approximation }} \cr
& {\text{Then using Newton's method we have}}: \cr
& {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 15 - \frac{{{{\left( {15} \right)}^2} - 250}}{{2\left( {15} \right)}} = 15.8333 \cr
& {c_3} = {c_2} - \frac{{f\left( {{c_2}} \right)}}{{f'\left( {{c_2}} \right)}} = 15.8333 - \frac{{{{\left( {15.8333} \right)}^2} - 250}}{{2\left( {15.8333} \right)}} = 15.8114 \cr
& {\text{In the same way}} \cr
& {c_4} = 15.8114 - \frac{{{{\left( {15.8114} \right)}^2} - 250}}{{2\left( {15.8114} \right)}} = 15.8113 \cr
& {\text{Since }}{c_3} = {c_4} = 15.811,{\text{ to the nearest thousand}}{\text{, }}\sqrt {250} = 15.811 \cr} $$
