Question 1082018
The formula we'll use is
{{{n = ( (z*s)/E )^2}}}


where
n = minimum sample size
z = standard normal critical value
s = sample standard deviation, which approximates the population standard deviation {{{sigma}}} (sigma)
E = margin of error


We're given
s = 2.3
E = 0.1


At 95% confidence, the critical value is
z = 1.96
which is approximate. 
This value is found using a calculator or a table.


Let's plug s = 2.3, E = 0.1 and z = 1.96 into the formula to get


{{{n = ( (z*s)/E )^2}}}


{{{n = ( (1.96*2.3)/0.1 )^2}}}


{{{n = ( (4.508)/0.1 )^2}}}


{{{n = ( 45.08 )^2}}}


{{{n = 2032.2064}}}


{{{n = 2033}}} Round up to the nearest whole number


We round up to ensure we clear the hurdle. If n = 2032 then the sample size isn't large enough to make E = 0.1


Note how plugging n = 2032 into the margin of error formula {{{E = z*(s/sqrt(n))}}} yields
{{{E = z*(s/sqrt(n))}}}
{{{E = 1.96*(2.3/sqrt(2032))}}}
{{{E = 0.1000050786112}}}
The margin of error (E) is larger than 0.1 so this sample size isn't large enough.


While plugging n = 2033 into the margin of error formula yields
{{{E = z*(s/sqrt(n))}}}
{{{E = 1.96*(2.3/sqrt(2033))}}}
{{{E = 0.09998048014111}}}
telling us we have cleared the threshold needed. We now have E < 0.1 making this the smallest sample size possible. Anything larger for n and E simply gets smaller. 


note: "margin of error within 0.1" means that the error is either less than 0.1 or at 0.1 exactly. 

-----------------------------------------------------------------
-----------------------------------------------------------------


Answer: The minimum sample size needed is <font color=red>2033</font>