Answer
The specific heat of the substance is $341~J/kg~C^{\circ}$
Work Step by Step
The energy $Q$ required to raise the temperature of a substance is:
$Q = mc~\Delta T$
We can find the heat energy gained by the aluminum, water, and glass.
$Q = m_a~c_a~\Delta T+ m_w~c_w~\Delta T+m_w~c_w~\Delta T$
$Q = \Delta T~(m_a~c_a+ m_w~c_w+m_w~c_w)$
$Q = (35.0^{\circ}C-10.5^{\circ}C)~[(0.105~kg)(900~J/kg~C^{\circ})+(0.185~kg)(4186~J/kg~C^{\circ})+(0.017~kg)(840~J/kg~C^{\circ})]$
$Q = 21,638~J$
The heat energy lost by the substance will be equal in magnitude to the heat energy gained by the aluminum, water, and the glass. We can find the specific heat $c_s$ of the substance.
$Q = m_s~c_s~\Delta T$
$c_s = \frac{Q}{m_s~\Delta T}$
$c_s = \frac{21,638~J}{(0.215~kg)(330^{\circ}C-35.0^{\circ}C)}$
$c_s = 341~J/kg~C^{\circ}$
The specific heat of the substance is thus $341~J/kg~C^{\circ}$.